Suppose we have a vector field $f(x, y) = (3, x + y)$ and a curve $C$ that is parameterized by $\alpha(t) = (t, 3t + 3)$ for $0 < t < 1$. What is the line integral of $f$ along $C$ ? $ \int_C f \cdot d\alpha = $
Explanation: Given a vector field $f$, a parameterization $\alpha$, and bounds $t_0$ and $t_1$, we can calculate the line integral as follows: $ \int_C f \cdot d\alpha = \int_{t_1}^{t_2} f(\alpha(t)) \cdot \alpha'(t) \, dt$ Here, $f(x, y) = (3, x + y)$ and $\alpha(t) = (t, 3t + 3)$. $\begin{aligned} &f(\alpha(t)) = (3, t + 3t + 3) = (3, 4t + 3) \\ \\ &\alpha'(t) = (1, 3) \end{aligned}$ Now we can rewrite our line integral as a single-variable integral. $ \int_C f \cdot d\alpha = \int_0^1 (3, 4t + 3) \cdot (1, 3) \, dt$ Let's solve the integral. $\begin{aligned} \int_0^1 (3, 4t + 3) \cdot (1, 3) \, dt &= \int_0^1 3 + 12t + 9 \, dt \\ \\ &= \int_0^1 12t + 12 \, dt \\ \\ &= \left[ 6t^2 + 12t \right]_0^1 \\ \\ &= 6 + 12 \\ \\ &= 18 \end{aligned}$ In conclusion, the line integral $ \int_C f \cdot d\alpha = 18$.